# How do you write # f(x)= x^2 + 6x + 12# in vertex form?

##### 2 Answers

#### Explanation:

The equation of a parabola in

#color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where " (h,k)" are the coordinates of the vertex"# and a is a constant.

#"Rearrange " f(x)=x^2+6x+12" into this form"#

#"using the method of "color(blue)" completing the square"#

#f(x)=(x^2+6xcolor(red)(+9))color(red)(-9)+12#

#color(white)(f(x))=(x+3)^2+3larrcolor(red)" in vertex form"#

#### Explanation:

The general vertex form is

with vertex at

Given

Extracting the "spread" factor

Completing the square

Re-writing with a squared binomial and some simplification:

...or in explicit vertex form:

(note that I dropped the

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In order to check that I didn't make any mechanical errors, I've plotted the original equation, and it does appear to have the vertex indicated by the derived equation: